Derivation of Legendre's Duplication Formula
Formula
$$ \Gamma (2r) = {{2^{ 2r - 1} } \over { \sqrt{ \pi } } } \Gamma \left( r \right) \Gamma \left( {{1} \over {2}} + r \right) $$
Explanation
The way it splits is not that pretty, but being able to break the argument into smaller pieces is certainly a useful fact. The derivation itself is not particularly difficult if we use a lemma derived from the beta function.
Derivation
$$ B(p,q) = {{\Gamma (p) \Gamma (q)} \over {\Gamma (p+q) }} = \int_{0}^{1} t^{p-1} (1-t)^{q-1} dt $$ Letting $r:= p=q$ for this, we get $$ {{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} = \int_{0}^{1} t^{r-1} (1-t)^{r-1} dt $$ Substituting $\displaystyle t = {{1+s} \over {2}}$, since $\lambda (s) := \left( 1 - s^2 \right)^{r-1}$ is an even function, $$ \begin{align*} {{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} =& {{1} \over {2}} \int_{-1}^{1} \left( {{1+s} \over {2}} \right)^{r-1} \left( {{1-s} \over {2}} \right)^{r-1} ds \\ =& {{1} \over {2^{1 + 2(r-1)} }} \int_{-1}^{1} \left( 1 - s^2 \right)^{r-1} ds \\ =& 2^{1 - 2r} \cdot 2 \int_{0}^{1} \left( 1 - s^2 \right)^{r-1} ds \end{align*} $$
Corollary of the trigonometric representation of the beta function: $$ B(x,y) = 2 \int_{0}^{1} t^{2x-1} \left( 1 - t^2 \right)^{y-1} dt $$
Substituting $\displaystyle x = {{1} \over {2}}$ and $y = r$ into the formula above, $$ B \left( {{1} \over {2}} , r \right) = 2 \int_{0}^{1} \left( 1 - t^2 \right)^{r-1} dt $$ so $$ {{\Gamma (r) \Gamma (r)} \over {\Gamma (2r) }} = 2^{1 - 2r} B \left( {{1} \over {2}} , r \right) = 2^{1 - 2r} {{\Gamma \left( {{1} \over {2}} \right) \Gamma (r)} \over {\Gamma \left( {{1} \over {2}} + r \right) }} $$ is obtained. Since $\displaystyle \Gamma \left( {1 \over 2} \right) = \sqrt{\pi}$ from the reflection formula, $$ {{\Gamma (r)} \over {\Gamma (2r) }} = 2^{1 - 2r} {{\sqrt{\pi} } \over {\Gamma \left( {{1} \over {2}} + r \right) }} $$ Rearranging with respect to $\Gamma (2r)$, $$ \Gamma (2r) = {{2^{2r-1} } \over { \sqrt{ \pi } } } \Gamma \left( r \right) \Gamma \left( {{1} \over {2}} + r \right) $$
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